All categories

3 years ago

Use the diagram below modeling a football kicked from a horizontal surface B

Physics

2 answers:

Bas_tet [7]3 years ago

7 0

Answer:

<h3>B is right try that that's hard tho lol:) </h3>

djyliett [7]3 years ago

5 0

B is the correct one

You might be interested in

Who else got stuck on a question and is NOT ready for test return at 2? its 1:58 AAAAAAA

k0ka [10]

Good luckkkk hope you do well

3 0

3 years ago

By looking at the relative positions of the elements calcium, Ca, fluorine, F, sulfur, S, and oxygen, O, in the Periodic Table,

SVEN [57.7K]

Answer:

id say the first option.

Explanation:

hope this helps you!

8 0

3 years ago

For those who think electric cars are sissy, Keio University in Japan has tested a 22-ft long prototype whose eight electric mot

Kruka [31]

Answer:

The percent of power extended against sea-level air drag is 50.15%

Explanation:

Length L is 22 ft, Power P is 590 hp,

Speed of air V = 175 mi/h. To convert to ft, we will multiply by 1.46

V = 175 × 1.46 = 255.5 ft

Frontal Area A is given as 23.5 ft

Coefficient of drag, Cd is 0.35

From text, Density of Air at sea level ρ = 0.00237 slug/ft³

The formula for calculating Drag Force F is

F = Cd × $\frac{1}{2}$ × ρAV²

Substitute the values above into the formula

F = 0.35 × $\frac{1}{2}$ × 0.00237 × 23.5 × 255.5²

= 0.35 × $\frac{1}{2}$ × 3635.78

= 636.26 lbf

But Power P = FV

Therefore we substitute for F and V to get P

P = 636.26 × 255.5

= 162564.43 ft.lbf/s

To convert to horsepower, we multiply by 0.00182 hp

= 162564.43 × 0.00182

= 295.87 hp

The percent of power extended against sea-level air drag is

= (295.87/590) × 100

= 50.15%

6 0

3 years ago

Read 2 more answers

Which statement is true? A) The work done to lift an object 6 meters is greater than the gravitational potential energy it gains

inna [77]

The work required to raise an object to a height is equal to the gravitational potential energy the object gains. <em>(C)</em>

4 0

4 years ago

Two uniformly charged conducting plates are parallel to each other. They each have area A. Plate #1 has a positive charge Q whil

Marrrta [24]

Answer:5

Explanation:

Given

First Plate has a charge of +Q and area A

Second Plate has a charge of -3 Q and area A

We Know electric Field due to sheet charge is given by

$E=\frac{Q}{2A\epsilon }$

Where Q=charge over the Plate

A=Area of plate

$\epsilon$ = Permittivity of free space

Electric Field Due to Positive charge will always be away from it while for negative charge it is towards it.

Net Electric Field at a point between between the Plates is the superimposition of two electric with direction

$E_{net}=\frac{Q}{2A\epsilon }+\frac{3Q}{2A\epsilon }$

$E_{net}=\frac{2Q}{A\epsilon }$

Net electric Field is towards the negative charged plate

5 0

4 years ago